Problem 1 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

A solution of two components containing \(n_{1}\) moles of the \(1^{\text {st }}\) component and \(n_{2}\) moles of the \(2^{\text {nd }}\) component is prepared. \(\mathrm{M}_{1}\) and \(\mathrm{M}_{2}\) are the molecular weights of component 1 and 2 respectively. If \(d\) is the density of the solution in \(g \mathrm{~mL}^{-1}, \mathrm{C}_{2}\) is the molarity and \(x_{2}\) is the mole fraction of the \(2^{\text {nd }}\) component, then \(\mathrm{C}_{2}\) can be expressed as: (a) \(\mathrm{C}_{2}=\frac{1000 x_{2}}{\mathrm{M}_{1}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\) (b) \(\mathrm{C}_{2}=\frac{d x_{2}}{\mathrm{M}_{2}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\) (c) \(\mathrm{C}_{2}=\frac{1000 d x_{2}}{\mathrm{M}_{1}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\) (d) \(\mathrm{C}_{2}=\frac{d x_{1}}{\mathrm{M}_{2}+x_{2}\left(\mathrm{M}_{2}-\mathrm{M}_{1}\right)}\)

Step-by-Step Solution

Answer

The correct expression for \( C_2 \) is \( \frac{1000 d x_2}{M_1 + x_2(M_2 - M_1)} \), option (c).

Step 1: Understand Molarity Formula

Molarity \(C\) is the number of moles of a solute (\(n_2\) in this case) divided by the volume of the solution in liters. Mathematically, it is expressed as \(C = \frac{n_2}{V}\). Our goal is to express \(C_2\), the molarity of the second component, using the given variables.

Step 2: Calculate Mole Fraction of Second Component

The mole fraction \(x_2\) is given by \(x_2 = \frac{n_2}{n_1 + n_2}\). This describes the fraction of moles of the second component as part of the total moles in the solution.

Step 3: Express Moles in Terms of Mole Fraction and Total Moles

From the definition of \(x_2\), rearrange to find \(n_2 = x_2 (n_1 + n_2)\). The mole fraction can help us link the number of moles of the solute to other properties in the solution.

Step 4: Determine Total Mass of the Solution

The total mass of the solution \(m\) is \(m = n_1 \cdot M_1 + n_2 \cdot M_2\). This uses the moles of each component multiplied by their respective molecular weights.

Step 5: Convert Mass to Volume using Density

Using the density \(d\), the volume \(V\) of the solution in liters can be expressed as \(V = \frac{m}{d \times 1000}\), since \(1 \text{ liter} = 1000 \text{ mL}\). Substitute the expression for mass \(m\).

Step 6: Substitute into Molarity Expression

Substitute both \(n_2 = x_2(n_1 + n_2)\) and \(V = \frac{(n_1 \cdot M_1 + n_2 \cdot M_2)}{d \times 1000}\) into the molarity formula \(C = \frac{n_2}{V}\) to find \(C_2\).

Step 7: Simplify Expression for Molarity

By working through the algebra, multiply the equations together: \[ C_2 = \frac{x_2 (n_1 + n_2) \cdot d \times 1000}{n_1 \cdot M_1 + n_2 \cdot M_2} \] Simplify further using the relation of \(x_2\) leading to the expression: \[ C_2 = \frac{1000 d x_2}{M_1 + x_2(M_2 - M_1)} \]

Step 8: Identify the Correct Expression

The simplified expression from the previous step matches option (c) from the given choices: \(C_2 = \frac{1000 d x_2}{M_1 + x_2(M_2 - M_1)}\).

Key Concepts

Mole Fraction

Mole fraction is a way to express the concentration of a component within a mixture. It represents the proportion of one type of molecule compared to the total number of molecules present. The formula for the mole fraction of a component is given by:\[x_2 = \frac{n_2}{n_1 + n_2}\]Here, \(n_2\) is the number of moles of the component of interest, and \(n_1\) is the number of moles of the other component in the solution.

• The value of \(x_2\) will always be between 0 and 1.
• If \(x_2\) equals 1, it means that component 2 is the only substance present.
• If \(x_2\) is 0, component 2 is not present at all.

Mole fraction is a unitless quantity because it is a ratio of similar quantities. It is especially useful when dealing with solutions and gases, as it provides a clear picture of how much of each component is present relative to the whole.
Understanding mole fraction helps in chemical calculations, such as determining the partial pressures of gases in a mixture, or as in this exercise, relating it to molarity to find other solution properties.

Density of Solution

The density of a solution gives us information about how much mass is packed into a given volume. It is defined as the mass of the solution divided by its volume:\[d = \frac{m}{V}\]where \(d\) is the density, \(m\) is the mass of the solution, and \(V\) is its volume. Here, we often measure density in terms of grams per milliliter (g/mL).

• Density helps in converting between mass and volume, which is crucial for finding molarity.
• The higher the density, the more mass is contained in a given volume.
• In this exercise, density allows us to determine the volume of the solution needed for molarity calculations by using: \(V = \frac{m}{d \times 1000}\).

Since density links mass to volume, it becomes a powerful tool in solutions to translate measurements into useful quantities. In practice, knowing the density of a solution can be critical for correctly preparing chemical solutions in a laboratory setting.

Molecular Weight

Molecular weight (or molar mass) is the mass of one mole of a substance expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a molecule. For example, water (\(H_2O\)) has a molecular weight calculated by adding the atomic masses of 2 hydrogen atoms and 1 oxygen atom.

• Molecular weight is vital for converting between moles and grams.
• Knowing the molecular weight helps in calculating the total mass of solutions and determining concentrations.
• In our example, \(M_1\) and \(M_2\) refer to the molecular weights of the two components in the solution.

By knowing the molecular weight, we can find out how much of each substance is present in terms of mass. This becomes essential when combining the mass and mole concepts to determine solution concentrations, like molarity, which requires both the molecular weight and the number of moles for calculation. Molecular weight thus bridges the gap between the atomic scale (moles) and the macroscopic scale (grams) in chemistry.